Chem 20 Final Review
Chemical Bonding
Bonding Theory: Valence Electrons & Orbitals
• According to bonding theory, valence electrons
are classified in terms of orbital occupancy.
(0 = empty, 1 = half filled , 2 = full)
– An atom with a valence orbital that has a single
electron can theoretically share that electron with
another atom
• Such an electron is called a BONDING ELECTRON
– An atom with a full valence orbital (2 e-’s), repels
nearby orbitals and wants to be alone
• Such a pairing is called a LONE PAIR
Covalent Bonding
 Both atoms have a high EN so neither atom “wins”
 The simultaneous attraction of two nuclei for a shared pair
of bonding electrons = covalent bond
Cl2 = diatomic
 EN difference can be zero = Cl – Cl
EN = 3.2 EN = 3.2
 EN difference can be small = H - Cl
EN = 2.2 EN = 3.2
 This is called a polar covalent bond – because
one side pulls on the electrons more but we
will learn more about this in Section 3.3
Ionic Bonding
• The EN of the two atoms are quite different
• The atom with the higher EN will remove the
bonding e- from the other atom
• Electron transfer occurs
– Positive and negative ions are formed which
electrically attract each other
EN = 0.9
EN = 3.2
Metallic Bonding
 Both atoms have a relatively low EN so atoms share valence
electrons, but no actual chemical reaction takes place
 In metallic bonding:
a) e-’s are not held very strongly by their atoms
b) the atoms have vacant valence orbitals
- This means the electrons are free to move around between
the atoms and the (+) nuclei on either side will attract them
Analogy: The positive nuclei are held
together by a glue of negative e-’s
Molecular Elements
• Many molecular elements are
diatomic and some are polyatomic
– You will need to memorize the formulas
of the 9 molecular elements as they will
not be given to you:
Name
Symbol
hydrogen
H2(g)
nitrogen
N2(g)
oxygen
O2(g)
fluorine
F2(g)
chlorine
Cl2(g)
iodine
I2(g)
bromine
Br2(g)
phosphorous
P4(g)
sulfur
S8(g)
Determining Lewis Formulas
• So why do we care about bonding capacity?
– If we know how many bonding e-’s an atom has, we
can predict what structure a molecular compound
willAtom
have
Number of
Number of
Bonding capacity
valence electrons bonding electrons
H
carbon
4
4
4
nitrogen
5
3
3
oxygen
6
2
2
halogens
7
1
1
hydrogen
1
1
1
I.e. Carbon can form 4 single bonds, 2 double bonds, 1 triple and 1 single, or 1 double and 2 singles
Lewis Formulas – Guided Ex. #2
Determine the Lewis formula & structural formula for
the nitrate ion, NO31. Count the valence electrons (*look for a net charge if an ion).
nitrogen = 1 x 5 valence e-’s = 5
oxygen = 3 x 6 valence e-’s = 18
23 + 1 (b/c net charge is -1) = 24
2. Which is the central atom? Nitrogen (in lesser quantity)
3. Arrange peripheral atoms around central atom and
place 1 pair of valence e-’s between them
N
Lewis Formulas – Guided Ex. #2
4. Place lone pairs on all peripheral atoms
to complete their octet
N
5. Place any remaining valence e-’s on the
central atom as lone pairs.
6. If the central atom’s octet is not
complete, move a lone pair from a
peripheral atom to a new position
between the peripheral and central
atom.
7. If the entity is a polyatomic ion, place
square brackets around the entire Lewis
formula and then write the net charge
outside the bracket on the upper right.
N
N
Pg. 95
Polarity
Chemists believe that molecules are made up of charged particles
(electrons and nuclei).
A polar molecule is one in which the negative
(electron) charge is not distributed symmetrically
among the atoms making up the molecule.
 Thus, it will have partial positive and negative charges on opposite sides
of the molecule.
A molecule with symmetrical electron distribution is a nonpolar
molecule.
 The existence of polar molecules can be demonstrated by running a
stream of water past a charged object.
 Demo: See Figure 9
PREDICTING AND EXPLAINING POLARITY
• Pauling explained the polarity of a covalent bond as
the difference in electronegativity of the bonded
atoms.
Cl2(g)
– If the bonded atoms have the same electronegativity,
they will attract any shared electrons equally and form a
nonpolar covalent bond.
– If the atoms have different electronegativities, they will
form a polar covalent bond.
– The greater the electronegativity difference, the more
polar the bond will be.
• For a very large electronegativity difference, the
difference in attraction may transfer one or more
electrons resulting in ionic bonding.
We use the
Greek symbol
delta to show
partial charges
Guided Practice #1
Go to Learning
Tip pg. 102
• Predict the polarity of the water molecule.
1) Draw the Lewis formula
O
2) VSEPR: Draw the stereochemical formula
H
H
Angular (bent)
3) Assign the EN of the atoms, assign δ– and δ+ to the bonds
4) •Draw
in the
bond
dipoles
The bond
dipoles
(vectors)
do not balance.
• Instead, they add together to produce a nonzero
molecular dipole (shown in red).
• This results in a polar molecule (explains bending water)
BACKGROUND
• There are three types of forces in matter:
1) Intranuclear force (bond) – bonds within the nucleus between protons and neutrons
(very strong)
2) Intramolecular force (bond) – bonds between atoms within the molecule or between
ions within the crystal lattice (quite strong)
3) Intermolecular force (bond) – bonds between molecules (quite weak); are
electrostatic (involve positive and negative charges)
There are 3 types of intermolecular bonds:
Weakest
a) Dipole-Dipole Forces (a.k.a. Polar Forces)
Medium
b) London Force (a.k.a. London Dispersion Force, Dispersion Force)
Strongest c) Hydrogen
Bonding
Note: “Van der Walls force” – includes London and dipole-dipole forces
1) Dipole-Dipole Force
• The simultaneous attraction between oppositely
charged ends of polar molecules.
▫ Simply put, the attraction between diploes
Dipole: a partial separation of positive and negative charges
within a molecule, due to electronegativity differences
▫ Dipole-dipole forces are among the weakest intermolecular
forces, but still control important properties (i.e. Solubility
because water is polar))
2) London Force
• Simultaneous attraction between a momentary dipole in a molecule
and the momentary dipoles in surrounding molecules
momentary dipole: an uneven distribution of electrons around a molecule,
resulting in a temporary charge difference between its ends
They last for just
the instant that the
electrons are not
distributed
perfectly even.
3) Hydrogen Bonding
• Occurs when a hydrogen atom bonded to a strongly
electronegative atom, (N, O and F) is attracted to a lone
pair of electrons in an adjacent molecule.
▫ Hydrogen nucleus (proton) is simultaneously attracted to two
pairs of electrons; one closer (in the same molecule) and one
further away (on the next molecule)
Why do you need a strongly
electronegative atom?
It pulls the hydrogen’s
electron away making it
“unshielded”, so the lone pair
on the other side can come
much closer
Gases
Comparing Kelvin and
Celsius Scales
• To convert degrees Celsius to Kelvin ,
you add 273.
K = °C + 273
• To convert Kelvin to degrees Celsius,
you subtract 273.
°C = K - 273
• Examples:
– What is 254 K in °C ?
-19°C
– What is -34°C in K ?
239K
Atmospheric Pressure
• Pressure exerted by air on all objects
Standard Temperature and Pressure (STP)
= 101.325 kPa and 0°C
•
•
But laboratory temperatures are not at 0°C
So scientists agreed on another set of conditions...
Standard Ambient Temperature and Pressure (SATP)
= 100 kPa and 25°C
•
Much closer to lab conditions – so scientists don’t freeze
Molar mass
• Now you try it…
• What is the molar mass of methanol CH3OH?
C = 12.01 g/mol x 1 = 12.01 g/mol
H = 1.01g/mol x 4 = 4.04 g/mol
O = 16.00g/mol x 1 = 16.00 g/mol
Molar mass of methanol = 32.05 g/mol
Gas Laws
• They are based on the temperature, pressure and
volume relationships that all gases have in common
1. Boyle’s Law P1V1 = P2V2
2. Charles’ Law
V1 = V2
T1 T2
3. Combined Gas Law P1V1 =P2V2
T1
T2
Example: Law of Combining Volumes
Use the law of combining volumes to predict the volume of oxygen required for the
complete combustion of 120 mL of butane gas from a lighter.
1) The first step is to write the balanced chemical equation, including what you are
given and what you need to find:
2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(g)
120 mL
V=?
2) From this chemical equation you can see that 13 mol of oxygen is required for
every 2 mol of butane. Therefore, the volume of oxygen has to be greater than
120mL by a factor of 13/2.
VO2: 120 ml C4H10 x ( 13 mL O2)
2 mL C4H10
= 780 mL
To make sure that the ratio is used in the correct order, you could include the
chemical formula with each quantity as shown above. Note the cancellation of the
units and chemical formulas
Molar Volume
STP = 22.4L/mol
SATP = 24.8 L/mol
• Molar volume is the same for all gases at the same
temperature and pressure (remember, all gases have the
same physical properties)
– At STP, molar volume = 22.4 L/mol (101.325 kPa and 0°C)
– At SATP, molar volume = 24.8 L/mol (100 kPa and 25°C)
• This can be used as a conversion factor just like molar mass!
At STP, one mole of gas has a volume of
22.4 L, which is approximately the
volume of 11 “empty” 2 L pop bottles.
Molar Volume – Practice
4.
A propane tank for a barbecue contains liquefied propane. IF
the tank mass drops by 9.1 kg after a month’s use, what
volume of propane gas at SATP was used for cooking?
Molar mass (M): C3H8(g)
= 44.11 g/mol
V C3H8(g) : 9.1 kg x (1 mol ) x ( 24.8 L) = 5.1 kL
44.11g
1 mol

What if I wanted your answer in litres?
5.1 kL x 1000L
1 kL
= 5100 L = 5.1 x 103 L
STP = 22.4L/mol
SATP = 24.8 L/mol
IDEAL GAS LAW
• Before getting too far into this law, it is important to understand
the difference between an ideal gas and a real gas....
– IDEAL GAS – does not really exist, it is hypothetical
• Follows all gas laws perfectly under all conditions
• Does not condense when cooled
• Assumes that the particles have no volume and are not attracted
to each other
– REAL GAS – does not follow gas laws exactly, it deviates at
low temperatures and high pressures
• Condenses to liquid or sometimes solid when cooled or under pressure
• Particles are attracted to each other and have volume
• Behaves like an ideal gas at higher temperatures and lower pressures
Using the Ideal Gas Law
Example One: What mass of neon gas should be
introduced into an evacuated 0.88L tube to
produce a pressure of 90 kPa at 30°C?
PV = nRT
n = PV
RT
n =
(90kPa)(0.88L)
(8.314 L•kPa/mol•K)(303K)
n = 0.0314 mol x 20.18 g = 0.63 g
1 mol
Using the Ideal Gas Law – Practice
3. Predict the volume occupied by 0.78 g of
hydrogen at 22°C and 125 kPa
n H2 = 0.78 g x 1 mol = 0.386 mol
2.02g
PV = nRT 
V = nRT
P
V = (0.386 mol)(8.314L•kPa/mol•K)(295K)
125 kPa
V = 7.573 L
V = 7.6 L
Solutions
Dissociation vs. Ionization
• What is the difference between dissociation and
ionization?
• Both produce (aq) ions...
• Dissociation, however, is the separation of ions
that already exist before dissolving in water
M+X- (s)  M+ (aq) + X-(aq)
• Ionization involves the production of new ions,
specifically hydrogen ions
HX0 (aq)  H+ (aq) + X-(aq)
SUMMARY
Substance
Process
Molecular
Disperse as individual,
neutral molecules
Ionic
Dissociate into individual
ions
Base (ionic hydroxide)
Dissociate into positive ions
and hydroxide ions
Acid
Ionize to form hydrogen
ions and anions
Reference pg. 201
General Equation
XY (s/l/g)  XY (aq)
MX (s)  M+ (aq) + X-(aq)
MOH (s)  M+ (aq) + OH-(aq)
HX (s/l/g)  H+ (aq) + X-(aq)
Bicarbonates: 320 PPM
Calcium Ion: 150 PPM
Fluoride: 0.12 PPM
Magnesium: 4.2 PPM
Potassium: 1.2 PPM
Sulfates: 43 PPM
Sodium: 11 PPM
• Amount Concentration (aka Molar
Concentration) (mol/L) – the amount of
moles of solute per litre of solution; this is the
most common form of concentration used in
chemistry
• c = n solute (mol)
Vsolution (L)
• Al2(SO4)3 (aq)  2 Al 3+(aq) + 3 SO42- (aq)
• c = 0.30 mol/L
• [Al 3+(aq)] = 0.30 mol/L x (2)
mol/L
1
= 0.60
• [SO42- (aq) ] = 0.30 mol/L x (3) = 0.90
Determining the mass of pure solid for a
Standard Solution
• Use conversion factors to determine the values for both the amount in
moles and the mass of solid required.
• Because you are working with one substance, you do not need a balanced
equation (No need for a mol ratio)
• Volume of the solution and its molar concentration are needed.
• Example: To prepare 250.0mL of 0.100 mol/L solution of sodium carbonate,
the mass needed is:
0.2500 L x 0.100 mol x 105.99 g = 2.65 g
1L
1 mol
Determining the volume of Stock Solution for
a Standard Solution
• We know that the number of moles does not change
when diluting a solution, but the concentration and
volume will. So we will use the dilution formula:
C1V1 = C2V2
• To use the dilution formula, you must make sure the units are
consistent for both the c and v (i.e. both in mL or both in L)
• Example: How would you prepare 100 mL of 0.40 mol/L MgSO4(aq)
from a solution of 2.0 mol/l MgSO4(aq)
C1V1 = C2V2
(2.0mol/L) (V1) = (0.40mol/L) (100mL)
V1 = 20 mL
Acids and Bases
Empirical Definitions
Acid – a substance which dissolves in water to produce a solution that:
–
–
–
–
–
Tastes sour
Turns blue litmus red
Conducts electricity
Reacts with active metals to produce H2(g)
Neutralizes Bases
Base – a substance which dissolves in water to produce a solution that:
–
–
–
–
Tastes bitter; feels slippery
Turns red litmus blue
Conducts electricity
Neutralizes acids
Theoretical Definitions
a) Arrhenius:
Acid – a substance that forms an acidic solution by dissolving in
water to produce free hydrogen ions (H+(aq)) in solution
–
Example: HCl (aq)  H+ (aq) + Cl- (aq)
Base – a substance that forms a basic solution by dissolving in
water to produce free hydroxide ions (OH-(aq)) in solution
–
Example: NaOH(aq)  Na+(aq) + OH–(aq)
Theoretical Definitions
b) Modified Definition:
Acid – a species that forms an acidic solution by reacting with water
to produce hydronium ions (H3O+(aq))
–
Example: HCl (aq) + H2O(aq) H3O+(aq) + Cl- (aq)
Base – a species that forms a basic solution by reacting with water
to produce hydroxide ions (OH-(aq))
–
Example: NH3 (aq) + H2O(aq) NH4+(aq) + OH–(aq)
The hydronium ion (hydrated proton) – was discovered by Paul
Giguère at the Université Laval in 1957.
Practice – Naming Acids
IUPAC
– HClO4(aq)
aqueous hydrogen perchlorate
– HClO(aq)
aqueous hydrogen hypochlorite
Traditional
– HNO2(aq)
nitrous acid
– HNO3(aq)
nitric acid
– H2SO4(aq)
aqueous hydrogen sulfate
sulfuric acid
– H2SO3(aq)
aqueous hydrogen sulfite
sulfurous acid
– H3PO4(aq)
aqueous hydrogen phosphate
phosphoric acid
pH – power of hydrogen
[H3O +(aq)] = 10 -pH
The pH scale is used to communicate a broad range of hydronium ion concentrations.
Most common acids and bases have pH values between 0 and 14
Summary
pH = -log [H3O+(aq)]
[H3O+(aq)] =10 –pH
pOH = -log [OH -(aq)]
[OH -(aq)] =10 –pOH
• The number of digits following the decimal point in a pH or pOH
value is equal to the number of significant digits in the
corresponding hydronium or hydroxide concentration.
• For both pH and pOH, an inverse relationship exist between the
ion concentration and the pH or pOH. The greater the hydronium
ion concentration, the lower the pH is.
Practice
• Pg. 242 #4-7 (pH)
• Pg. 243 #9-11 (pOH)
Acid – Base Indicators
• Other acid-base indicators:
– Because the chemical structure of each indicator is different, the pH at which the
indicator changes from the HIn(aq) form to the In-(aq) form is different for each
indicator. (See inside back cover of textbook)
Common name
Bromothymol blue
Phenolphthalein
Color of
HIn(aq)
pH range of colour
change
Color of In(aq)
Yellow
6.0-7.6
Blue
Colourless
8.2-10.0
Pink
Practice
• Lab Exercise 6.B pg. 247
– Complete the Analysis Portion of the lab below
Solution
After addition to samples of the
solution …
A
●
●
●
●
methyl violet was blue
methyl orange was yellow
methyl red was red
phenolphthalein was colourless
B
●
●
●
●
indigo carmine was blue
phenol red was yellow
bromocresol green was blue
methyl red was yellow
C
●
●
●
●
phenolphthalein was colourless
thymol blue was yellow
bromocresol green was yellow
methyl orange was orange
pH range implied by
each indicator
pH of solution
More Practice....
• HI(aq) – explain acidic properties
– HI(aq) + H2O(l)  H3O+(aq) + I- (aq)
• NaCH3COO(aq) – explain basic properties
– NaCH3COO(aq)  Na+ (aq) + CH3COO-(aq)
Simple Dissociation
– CH3COO-(aq) + H2O(l)  CH3COOH (aq) + OH-(aq)
• HOCl(aq) – explain acidic properties
– HOCl + H2O(l)  H3O+(aq) + OCl- (aq)
Try Three More...
• H3PO4(aq) – explain acidic properties
– H3PO4(aq) + H2O(l)  H3O+(aq) + H2PO4- (aq)
• Na2SO4 (aq) – explain basic properties
– Na2SO4 (aq)  2 Na+ (aq) + SO42-(aq)
Simple Dissociation
– SO42-(aq) + H2O(l)  HSO4- (aq) + OH-(aq)
• Sr(OH)2(aq) – explain basic properties
– Sr(OH)2(aq)  Sr2+(aq) + 2 OH- (aq)
Summary
• Acids are substances that react with water to
produce hydronium ions
• Most bases are substances that react with water
to produce hydroxide ions
• Neutralization can be explained as the reaction
between hydronium ions and hydroxide ions to
produce water.
• Try pg. 253 #4,5
The Difference: Using the
Modified Arrhenius Theory
• Strong Acids: have high conductivity, high rate of reaction w/
metals and carbonates and a relatively low pH
• These empirical properties suggest many ions are present (lots
of H3O+ ions present); which is consistent with the idea that
strong acids react completely (>99%) with water to form
hydronium ions
>99%
HCl(aq) + H2O(l)  H3O+(aq) + Cl-(aq)
The Difference: Using the
Modified Arrhenius Theory
• Weak Acids: have low conductivity, a lower rate of reaction
w/ active metals and carbonates and a relatively high pH
• These empirical properties suggest fewer hydronium ions
are present
• Based on this evidence, a weak acid reacts incompletely
(<50%) with water to form relatively few hydronium ions
<50%
CH3COOH(aq) + H2O(l)  H3O+(aq) + CH3COO-(aq)
Strong and Weak Bases
• Strong Bases – all soluble ionic hydroxides that
dissociate completely (>99%) to release
hydroxide ions
NaOH(s)  Na+(aq) + OH-(aq)
• Weak Bases – an ionic or molecular substance
that reacts partially (<50%) with water to
produce relatively few hydroxide ions
<50%
NH3 (aq) + H2O(aq) OH–(aq) + NH4+(aq)
Summary
Strong Acids
Empirical
properties
(need same
concentration
& temperature)
Modified
Arrhenius
Theory
Weak Acids
Strong Bases
Weak Bases
Very low pH
Med to low pH
Very high pH
Med to high pH
High
conductivity
Low
conductivity
High
conductivity
Low
conductivity*
Fast reaction
rate
Slow reaction
rate
Fast reaction
rate
Slow reaction
rate
Completely
react with
water to form
H3O+(aq) ions
Partially react
with water to
form H3O+(aq)
ions
Completely
react with
water to form
OH-(aq) ions
Partially react
with water to
form OH-(aq)
ions
* Applies only to weak bases that are molecular
Polyprotic Acids
• With all of these possible reactions you might think that phosphoric acid is very
acidic, but only the first reaction is significant.
• Also, the first reaction is only 50% complete, so it is actually a weak acid and it has a
pH noticeable higher than a strong acid at the same concentration.
Acid
Concentration
pH
HCl(aq)
0.1 mol/L
1.0
H3PO4(aq)
0.1 mol/L
1.7
• Based on this evidence:
• In general, polyprotic acids are weak acids whose reaction with water
decreases with each successive step.
NOTE: H2SO4(aq) is a notable exception to this rule. It is a strong acid because it’s
first reaction with water is essentially complete. However the second reaction is
much less than 50% complete.
Stoichiometry
What do you remember?
In Science 10 you learned about five reaction types, can you match them up
 Composition (Formation)  CH4(g) + O2(g)  CO2(g) + H2O(g)
 Decomposition
 Mg(s) + O2(g)  MgO(s)
 Combustion
 Cu(s) + AgNO3(aq)  Ag(s) + Cu(NO3)2(g)
 Single Replacement
 CaCl2(aq) + Na2CO3(aq)  CaCO3(s) +
NaCl(aq)
 Double Replacement
 H2O(l)  O2(g) + H2(g)
Using the solubility table:
When cancelling spectator ions, they
must be identical in every way:
chemical amount, form (atom, ion,
molecule) and state of matter
Practice
•
Write the net ionic equation for the reaction of aqueous barium chloride and
aqueous sodium sulfate. (Refer to the solubility table)
•
1) BaCl2(aq) + Na2SO4(aq)  BaSO4(s) + 2NaCl(aq)
•
2) Ba2+(aq) + 2Cl-(aq) +2Na+(aq) + SO42-(aq)  BaSO4(s) + 2Na+(aq) + 2Cl-(aq)
(Complete ionic equation)
•
3) Ba2+(aq) + 2Cl-(aq) +2Na+(aq) + SO42-(aq)  BaSO4(s) + 2Na+(aq) + 2Cl-(aq)
•
4) Ba2+(aq)) + SO42-(aq)  BaSO4(s) (Net ionic equation)
•
Ions that are present but do not take part in (change during) a reaction are called
spectator ions (like spectators at a sports game: they are present but do not take
part in the game)
Limiting and Excess Reagents
•
When no further changes appear to be occurring, we assume that all of the AgNO3(aq)
that was initially present has now been completely reacted.
•
A limiting reagent is the reactant whose entities are completely consumed in a
reaction, meaning the reaction stops.
– In order to make sure this happens, more of the other reactant must be present
than is required
•
An excess reagent is the reactant whose entities are present in surplus amounts, so
that some remain after the reaction ends..
•
In our reaction: much more copper was used than
needed (evidenced
by the unreacted copper) so we
assume the reaction ended when
no more silver ions
were left, so silver nitrate was the limiting
reagent.
Stoichiometry Calculations
(Measured quantity)
solids/liquids
m 
n
mole
ratio
solids/liquids
m 
(Required quantity)
n
Practice #3 (Mass Stoichiometry)
What mass of iron (III) oxide is required to
produce 100.0 g of iron?
Fe2O3(s) +
3 CO(g)
m=?
M = 159.70g/mol
m Fe2O3(s): 100.0 g x

2 Fe(s)
+
3 CO2(g)
m = 100.0g
M = 55.85 g/mol
1 mol x 1 mol x 159.70 g
55.85 g
2 mol
1 mol
= 143.0 g Fe2O3
Percent Yield for Reactions
• We can use stoichiometry to test experimental designs, technological skills,
purity of chemicals, etc. We evaluate these by calculating a percent yield.
– This is the ratio of the actual (experimental) quantity of product obtained and
the theoretical (predicted) quantity of product obtained from a stoichiometry
calculation
– Percent yield = actual yield
predicted yield
x 100
– Some forms of experimental uncertainties:
•
•
•
•
All measurements (limitations of equipment)
Purity of chemical used (80-99.9% purity)
Washing a precipitate (some mass is lost through filter paper)
Estimation of reaction completion (qualitative judgements i.e.
color)
Percent Yield Example #1
 Example: In a chemical analysis, 3.00 g of silver nitrate in
solution was reacted with excess sodium chromate to produced
2.81 g of filtered, dried precipitate.
 Predicted value: 2AgNO3(aq) + Na2CrO7(aq) Ag2CrO7(s) +2NaNO3(aq)
m = 3.00 g
M = 169.88g/mol
3.00g x
1 mol x
169.88g
1 mol x
2 mol
 Percent yield = actual yield
predicted yield
m=
M = 331.74 g/mol
331. 74 g = 2. 93 g
1 mol
x 100%
= 2.81g x 100%
2.93g
= 95.9%
Gas Stoichiometry
• If 300g of propane burns in a gas barbecue,
what volume of oxygen measured at SATP is
required for the reaction?
• Remember: 24.8L/mol for SATP
C3H8(g) +
5O2(g)  3CO2(g) + 4H2O(g)
m = 300g
44.11g/mol
300 g
x 1 mol
44.11 g
V=?
24.8L/mol
x 5 mol
1 mol
x
24.8 L = 843 L O2(g)
1 mol
**Remember – molar volume is the conversion factor for gases just like molar
mass is the conversion factor in gravimetric stoichiometry
Gas Stoichiometry
• What volume of ammonia at 450kPa and 80oC can be obtained from the
complete reaction of 7.5kg of hydrogen with nitrogen?
2N2(g) +
3H2(g) 
2NH3(g)
m = 7500g
V=?
M = 2.02 g/mol P = 450kPA
T = 353.13K
7500 g x 1 mol x
2.02 g
PV = nRT
 V = nRT
P
2 = 2475.2475 mol NH3(g)
3
= (2475.2475 mol)(8.314kpa•L/mol•K)(353.15K)
(450kPa)
= 16150.10L  1.6 x 104 L of NH3(g)
Solution Stoichiometry
 In an experiment, a 10.00 mL sample of sulfuric acid solution reacts
completely with 15.9 mL of 0.150 mol/L potassium hydroxide. Calculate
the amount concentration of the sulfuric acid.

H2SO4(aq) + 2KOH(aq)
V = 10.00mL
c=?

2H2O(l) + K2SO4(aq)
V = 15.9 mL
0.150 mol/L
15.9mL x 0.150 mol x 1mol
1L
2 mol
x
1
= 0.119 mol/L
10.0 mL
Chemical Analysis
Colorimetry
Ion
Solution colour
Group 1, 2, 17
Colourless
Cr2+(aq)
Blue
Cr3+(aq)
Green
Co2+(aq)
Pink
Cu+(aq)
Green
Cu2+(aq)
Blue
Fe2+(aq)
Pale-green
Fe3+(aq)
Yellow-brown
Mn2+(aq)
Pale pink
Ni2+(aq)
Green
CrO4
2-
(aq)
Yellow
Cr2O72-(aq)
Orange
MnO4-(aq)
Purple
Which solution is which?
potassium dichromate
sodium chloride
sodium chromate
potassium permanganate
nickel (II) nitrate
copper (II) sulfate
6
2
4
1
3
5
1) You want to test the stoichiometric method using the
reaction of 2.00 g of copper(II) sulfate in solution with an
excess of sodium hydroxide in solution. What would be a
reasonable mass of sodium hydroxide to use?
 To answer this question, you need to calculate the minimum
mass required and then add 10%.
CuSO4(aq) + 2 NaOH(aq)  Cu(OH)2(s) + Na2SO4(aq)
2.00g
m=?
159.62g/mol 40.00g/mol
2.00 g x
1 mol x 2
159.62 g
1
x
40.0g
1 mol
= 1.00 g
10% = 0.10g
1.00 g x 1.10%= 1.10g
Practice pg. 321 #2
2) If 10.0g of copper is placed in solution of 20.0g of silver
nitrate, which reagent will be the limiting reagent?
 All reactants must be converted to moles, then using the mole
ratio, determine which reactant will run out first.
Cu(s)
+ 2 AgNO3(aq)  2 Ag(s) + Cu(NO3)2(aq)
10.0g
given: 0.157 mol
20.0g
0.0589 mol
n Cu(s): 10.0g x 1 mol = 0.157 mol x 2 = 0.315 mol
63.55 g
1
n AgNO3: 20.0g x 1 mol = 0.118 mol x 1 = 0.0589 mol
169.88 g
2
You must test one of the values using the mole ratio.
Assume one chemical is completely used up and see
if enough of the second chemical is present.
That much silver
nitrate is not
available so
copper is not the
limiting reagent
More copper than
that is available so
silver nitrate is the
limiting reagent
Limiting and Excess Reagents
Summary
• Identify the limiting reagent by choosing either
reagent amount, and use the mole ratio to
compare the required amount with the amount
actually present.
• The quantity in excess is the difference between
the amount of excess reagent present and the
amount required for complete reaction.
• A reasonable reagent excess to use to ensure
complete reaction is 10%.
Chemical Analysis by Titration
• Titration – is a common
experimental design used to
determine the amount concentration
of substances in solution.
– The solution of known concentration
may be either the titrant or the
sample; it makes no difference to the
analysis
• Titration breakdown:
– Carefully adding a solution (titrant)
from a burette into a measured, fixed
volume of another solution (sample) in
an Erlenmeyer flask until the reaction
is judged to be complete
Chemical Analysis by Titration
• Burette – precisely marked glass cylinder with a
stopcock at one end. Allows precise, accurate
measurement and control of the volume of
reacting solution.
• When doing a titration, there will be a point at
which the reaction is complete; when chemically
equivalent amounts of reactants have combined.
This is called the equivalence point:
– Equivalence point – the point during a
titration at which the exact theoretical
chemical amount of titrant has been added to
the sample. (QUANTITATIVE)
• To measure this equivalence point experimentally,
we look for a sudden change in an observable
property, such as color, pH, or conductivity. This
is called the endpoint. (QUALITATIVE)
Chemical Analysis by Titration
•
•
•
•
An initial reading of the
burette is made before any
titrant is added to the sample.
Then the titrant is added until
the reaction is complete;
when a final drop of titrant
permanently changes the
colour of the sample.
The final burette reading is
then taken.
The difference between the
readings is the volume of
titrant added.
Near the endpoint, continuous gentle swirling of the solution is important
Sample Problem
• Determine the concentration of hydrochloric acid in a commercial solution.
• A 1.59g mass of sodium carbonate, Na2CO3(s), was dissolved to make 100.0mL of
solution. Samples (10.00mL) of this standard solution were then taken and
titrated with the hydrochloric acid solution.
• The titration evidence collected is below. Methyl orange indicator was used.
Trial
1
2
3
4
Final burette reading (mL)
13.3
26.0
38.8
13.4
Initial burette reading (mL)
0.2
13.3
26.0
0.6
Volume of HCl(aq) added
13.1
12.7
12.8
12.8
Indicator colour
Red
Orange Orange Orange
TIP: In titration analysis, the first trial is typically done very quickly. It is just for practice, to learn
what the endpoint looks like and to learn the approximate volume of titrant needed to get to the
endpoint. Greater care is taken with subsequent trials
•
The initial addition of the titrant (in the burette) to the acid does not produce large
changes. This relatively flat region of the pH curve is where a buffering action
occurs.
•
As the titration proceeds, and base is added, some of the acid is reacted with the
added base, but anywhere before the equivalence point some excess acid will
remain, so the pH stays relatively low.
• Very near the equivalence
point, a small excess of acid
becomes a small excess of
base with the addition of a
few more drops, so the pH
abruptly changes.
• The equivalence point is the
centre of this change, where
the curve is the most
vertical.
General
Rule
Strong Acid to
Weak Base:
pH
at equivalence
point is always
lower than 7
Strong Base to
Weak Acid: pH
at equivalence
point is always
higher than 7